This is the first problem from a 2005 International Math Olympiad page. The description here only partially matches the picture, but I'll explain.
I encourage you to draw a picture of the hexagon described in that problem statement, because I am starting here by drawing what you see above instead. I set point B of ΔABC at the origin on a coordinate plane and draw ray BA (A is not pictured yet) at a 60∘angle to the x-axis. Side BC will lie on the x-axis, and ∡B must be 60∘(same with ∡A and ∡C) because an equilateral triangle has 3 congruent angles. I've also marked a position for A1, which I've assigned the coordinates (x1,0).
Let's make the first side of the regular hexagon, which has equal side lengths and equal angle measures. Each angle measure is 6−2⋅1806, or 120∘. I can shoot a line at a 120∘angle toward the y-axis to form our first side, which I'll call s (it's ¯A1C2 in the picture).
ΔPBA1 is a 30-60-90 triangle, which tells us that the coordinates of point P are (0,√3x1). Line PA1, then, has the equation y=−√3x+√3x1, and the line that will form side AB of the triangle has the equation √3x. The intersection of these lines, point C2, is then at (x12,√3x12).
Finally, we can run a vertical line up from A1 to find C1. From C2, you can walk x12 to the right to get to the same horizontal position as A1 and up a distance of √3x12 to get to the same vertical position as P. This is the exact same walk we would make from point B to get to C2, so we know that we're still on the side of the triangle. And it's the same walk in terms of distances that one would make from point A1 to C2, so we know that C1C2 has length s. This means we also know the coordinates of C1: (x1,√3x1). It's probably also worth pointing out that x1=s. The equilateral triangle ΔA1BC2 shows us this.
Bridge with the Givens
Now we can use the givens to fill in the picture. And symmetry. That concept alone will save me paragraphs of writing. Since we can construct the right-hand side of this entire image using symmetry, I am able to easily give you the handy table listing the coordinates of all the points of the hexagon shown below.
| Point | Coordinates |
|---|---|
| A1 | (x1,0) |
| A2 | (x1+s,0) |
| B1 | (3x12+s,√3x12) |
| B2 | (x1+s,√3x1) |
| C1 | (x1,√3x1) |
| C2 | (x12,√3x12) |
What remains is for us to show that the lines A1B2, B1C2, and C1A2 are concurrent, or that they all meet at a point. And now that we have the coordinates of the vertices, we are left with the task of solving a system of 3 equations—or at least the task of showing that a system of 3 equations has a solution. I think we actually have too much information now, but that gives a lot of options.
Symmetry and Now Parallelisms FTW!
Since x1=s, a walk from point B to A1 and then up to C1 is the same as a walk from A1 to A2 and then up to B2. So, lines AB and A1B2 are parallel. Thus, the equation for line A1B2 is the same as the one for line AB (y=√3x) but just shifted to the right x1, or s, units: y=√3(x−x1).
Similarly, the equation for line C1A2 is the same as the one for line C2A1 (y=−√3x+√3x1) but shifted to the right x1, or s, units: y=−√3(x−x1)+√3x1.
Finally, the horizontal line B1C2 has the equation y=√3x12.
If you can slog through that algebra, the point of intersection of those three lines is at (3s2,√3s2). Whew!